4. \( \begin{aligned} 3 x+2 y & =12 \\ 2 x-4 y & =0\end{aligned} \)

To solve this system of equations, we can use the substitution or elimination method. Let's use the elimination method here. First, align both equations: 1. \( 3x + 2y = 12 \) (Equation 1) 2. \( 2x - 4y = 0 \) (Equation 2) To eliminate \(y\), we can multiply Equation 1 by 2 and Equation 2 by 1: - \( 2(3x + 2y) = 2(12) \) → \( 6x + 4y = 24 \) (Equation 3) - \( 1(2x - 4y) = 1(0) \) → \( 2x - 4y = 0 \) (Equation 2 remains unchanged) Now, add Equation 3 and Equation 2: \[ (6x + 4y) + (2x - 4y) = 24 + 0 \] \[ 6x + 2x + 4y - 4y = 24 \] \[ 8x = 24 \] \[ x = 3 \] Now substitute \(x = 3\) back into Equation 1 to find \(y\): \[ 3(3) + 2y = 12 \] \[ 9 + 2y = 12 \] \[ 2y = 3 \] \[ y = \frac{3}{2} \] The solution to the system of equations is \(x = 3\) and \(y = \frac{3}{2}\).